miniF2F

mathd_algebra_478

test

theorem mathd_algebra_478 (b h v : ℝ) (h₀ : 0 < b ∧ 0 < h ∧ 0 < v) (h₁ : v = 1 / 3 * (b * h)) (h₂ : b = 30) (h₃ : h = 13 / 2) : v = 65 := sorry

import Mathlib.Algebra.BigOperators.Basic import Mathlib.Data.Real.Basic import Mathlib.Data.Complex.Basic import Mathlib.Data.Nat.Log import Mathlib.Data.Complex.Exponential import Mathlib.NumberTheory.Divisors import Mathlib.Data.ZMod.Defs import Mathlib.Data.ZMod.Basic import Mathlib.Topology.Basic import Mathlib.Data.Nat.Digits open BigOperators open Real open Nat open Topology

The volume of a cone is given by the formula $V = \frac{1}{3}Bh$, where $B$ is the area of the base and $h$ is the height. The area of the base of a cone is 30 square units, and its height is 6.5 units. What is the number of cubic units in its volume? Show that it is 65.

We are given that $B = 30$ and $h = 6.5$ and asked to find $\frac{1}{3}Bh$. We find that \[\frac{1}{3}Bh = \frac{1}{3}(30)(6.5) = (10)(6.5) = 65.\]

numbertheory_4x3m7y3neq2003

test

theorem numbertheory_4x3m7y3neq2003 (x y : ℤ) : 4 * x^3 - 7 * y^3 ≠ 2003 := sorry

import Mathlib.Algebra.BigOperators.Basic import Mathlib.Data.Real.Basic import Mathlib.Data.Complex.Basic import Mathlib.Data.Nat.Log import Mathlib.Data.Complex.Exponential import Mathlib.NumberTheory.Divisors import Mathlib.Data.ZMod.Defs import Mathlib.Data.ZMod.Basic import Mathlib.Topology.Basic import Mathlib.Data.Nat.Digits open BigOperators open Real open Nat open Topology

Show that there are no integers $x$ and $y$ such that $4x^3 - 7y^3 = 2003$.

We have that $2003 = 1 + 7 \times 286$. So $4x^3 - 7y^3 = 2003 \iff 4x^3-1=7(y^3+286)$. We also observe that for every integer $x$, the rest of the division of $x^3$ by $7$ is $0$, $1$, or $6$. So the rest of the division of $4x^3-1$ by $7$ is $6$, $3$, or $2$. As a result, $7$ does not divide $4x^3-1$, but since $7(y^3+286)$ is divisible by $7$, the equation cannot have integers solutions.